Exercise 394Calculus › Single-variable calculus › Integration › Riemann integralL1Added 2 Nov 2026

The derivative of an integral with variable bounds

#fundamental theorem of calculus#chain rule#limits

Statement

Let $F(x)=\displaystyle\int_x^{x^2}e^{-t^2}\,dt$ . Compute $F'(x)$ .
Compute $\displaystyle\lim_{x\to0}\frac1x\int_0^x\cos(t^2)\,dt$ .

Hint

The function $e^{-t^2}$ has no elementary antiderivative, but you do not need one. Introduce $G(x)=\int_0^xe^{-t^2}dt$ and use $G'(x)=e^{-x^2}$ .

Show solutionHide solution

Solution

1. Let $G(u)=\int_0^ue^{-t^2}dt$ . Then $G'(u)=e^{-u^2}$ (fundamental theorem), and
$F(x)=G(x^2)-G(x).$
By the chain rule,
$F'(x)=2x\,G'(x^2)-G'(x)=2x\,e^{-x^4}-e^{-x^2}.$

2. Let $H(x)=\int_0^x\cos(t^2)\,dt$ . Then $H(0)=0$ and $H'(x)=\cos(x^2)$ , so
$\frac{H(x)}x=\frac{H(x)-H(0)}{x-0}\longrightarrow H'(0)=\cos0=1.$

Remark. The fundamental theorem says that *differentiation undoes integration*: $\frac d{dx}\int_a^xf=f(x)$ for continuous $f$ . The chain rule handles non-trivial bounds.

The derivative of an integral with variable bounds

Statement

Solution

More problems in Riemann integral